Thermodynamics of Ideal Gases

An ideal gas is a nice “laboratory” for understanding the thermodynamics of a ﬂuid with a non-trivial equation of state. In this section we shall recapitulate the conventional thermodynamics of an ideal gas with constant heat capacity. For more extensive treatments, see for example [67, 66].

D.1 Internal energy

In section 4.1 we analyzed Bernoulli’s model of a gas consisting of essentially 1 2 non-interacting point-like molecules, and found the pressure p = 3 ρ v where v is the root-mean-square average molecular speed. Using the ideal gas law (4-26) the total molecular kinetic energy contained in an amount M = ρV of the gas becomes, 1 3 3 Mv2 = pV = nRT , (D-1) 2 2 2 where n = M/Mmol is the number of moles in the gas. The derivation in section 4.1 shows that the factor 3 stems from the three independent translational degrees of freedom available to point-like molecules. The above formula thus expresses 1 that in a mole of a gas there is an internal kinetic energy 2 RT associated with each translational degree of freedom of the point-like molecules. Whereas monatomic gases like argon have spherical molecules and thus only the three translational degrees of freedom, diatomic gases like nitrogen and oxy-

gen have stick-like molecules with two extra rotational degrees of freedom or- thogonally to the bridge connecting the atoms, and multiatomic gases like carbon dioxide and methane have the three extra rotational degrees of freedom. Accord- ing to the equipartition theorem of statistical mechanics these degrees of freedom 1 will also carry a kinetic energy 2 RT per mole. Molecules also possess vibrational degrees of freedom that may become excited at high temperatures, but we shall disregard them here. The internal energy of n moles of an ideal gas is deﬁned to be,

k U = nRT , (D-2) 2

where k is the number of molecular degrees of freedom. A general result of thermodynamics (Helmholtz’ theorem [67, p. 154]) guarantees that for an ideal gas U cannot depend on the volume but only on the temperature. Physically a gas may dissociate or even ionize when heated, and thereby change its value of k, but we shall for simplicity assume that k is in fact constant with k = 3 for monatomic, k = 5 for diatomic, and k = 6 for multiatomic gases. For mixtures of gases the number of degrees of freedom is the molar average of the degrees of freedom of the pure components (see problem D.1).

D.2 Heat capacity

Suppose that we raise the temperature of the gas by δT without changing its volume. Since no work is performed, and since energy is conserved, the necessary amount of heat is δQ = δU = CV δT where the constant, k C = nR , (D-3) V 2 is naturally called the heat capacity at constant volume. If instead the pressure of the gas is kept constant while the temperature is raised by δT , we must also take into account that the volume expands by a certain amount δV and thereby performs work on the surroundings. The necessary amount of heat is now larger by this work, δQ = δU + pδV . Using the ideal gas law (4-26) we have for constant pressure pδV = δ(pV ) = nRδT . Consequently, the amount of heat which must be added per unit of increase in temperature at constant pressure is

Cp = CV + nR , (D-4)

called the heat capacity at constant pressure. It is always larger than CV because it includes the work of expansion.

The adiabatic index The dimensionless ratio of the heat capacities, C 2 γ = p = 1 + , (D-5) CV k is for reasons that will become clear in the following called the adiabatic index. It is customary to express the heat capacities in terms of γ rather than k, 1 γ C = nR , C = nR . (D-6) V γ − 1 p γ − 1 Given the adiabatic index, all thermodynamic quantities for n moles of an ideal gas are completely determined. The value of the adiabatic index is γ = 5/3 for monatomic gases, γ = 7/5 for diatomic gases, and γ = 4/3 for multiatomic gases.

D.3 Entropy

When neither the volume nor the pressure are kept constant, the heat that must be added to the system in an infinitesimal process is, δV δQ = δU + pδV = C δT + nRT . (D-7) V V It is a mathematical fact that there exists no function, Q(T,V ), which has this expression as differential (see problem D.2). It may on the other hand be directly verified (by insertion) that

δQ δT δV δS = = C + nR , (D-8) T V T V can be integrated to yield a function,

S = CV log T + nR log V + const , (D-9) called the entropy of the amount of ideal gas. Being an integral the entropy is only deﬁned up to an arbitrary constant. The entropy of the gas is, like its energy, an abstract quantity which cannot be directly measured. But since both quantities depend on the measurable thermodynamic quantities, ρ, p, and T , that characterize the state of the gas, we can calculate the value of energy and entropy in any state. But why bother to do so at all?

The two fundamental laws of thermodynamics The reason is that the two fundamental laws of thermodynamics are formulated in terms of the energy and the entropy. Both laws concern processes that may

take place in an isolated system which is not allowed to exchange heat with or perform work on the environment. The First Law states that the energy is unchanged under any process in an V isolated system. This implies that the energy of an open system can only change by exchange of heat or work with the environment. We actually used this law implicitly in deriving the heat capacities and the entropy.

...... The Second Law states that the entropy cannot decrease. In the real world, ...... V0 ...... the entropy of an isolated system must in fact grow. Only if all the processes ...... taking place in the system are completely reversible at all times, will the entropy stay constant. Reversibility is an ideal which can only be approached by very slow quasistatic processes, consisting of infinitely many infinitesimal reversible steps. Essentially all real-world processes are irreversible to some degree. A compartment of volume V0 inside an isolated box of volume V . Initially, Example D.3.1 (Joule’s expansion experiment): An isolated box of vol- the compartment contains ume V contains an ideal gas in a walled-off compartment of volume V0. When the an ideal gas with vacuum wall is opened, the gas expands into vacuum and fills the full volume V . The box in the remainder of the is completely isolated from the environment, and since the internal energy only de- box. When the wall breaks, the gas expands by itself pends on the temperature, it follows from the First Law that the temperature must to fill the whole box. The be the same before and after the event. The change in entropy then becomes reverse process would entail a decrease in entropy and ∆S = (CV log T + nR log V ) − (CV log T + nR log V0) = nR log(V/V0) never happens by itself. which is evidently positive (because V/V0 > 1). This result agrees with the Second James Prescott Joule (1818- Law, which thus appears to be unnecessary. 1889). English physicist. The strength of the Second Law becomes apparent when we ask the question Gifted experimenter who as of whether the air in the box could ever — perhaps by an unknown process to be the first demonstrated the discovered in the far future — by itself enter the compartment of volume V0, leaving equivalence of mechanical vacuum in the box around. Since such an event would entail a negative change in work and heat, a necessary step on the road to the First entropy which is forbidden by the Second Law, it never happens. Law. Demonstrated (in con- tinuation of earlier experi- ments by Gay-Lussac) that Isentropic processes the irreversible expansion of a gas into vacuum does not Any process in an open system which does not exchange heat with the environ- change its temperature. ment is said to be adiabatic. If the process is furthermore reversible, it follows that δQ = 0 in each infinitesimal step, so that the δS = δQ/T = 0. The entropy (D-9) must in other words stay constant in any reversible, adiabatic process. Such a process is for this reason called isentropic. By means of the adiabatic index (D-5) we may write the entropy (D-9) as,

¡ γ−1¢ S = CV log TV + const . (D-10) From this it follows that TV γ−1 = const , (D-11) for any isentropic process in an ideal gas. Using the ideal gas law to eliminate V ∼ T/p, this may be written equivalently as, T γ p1−γ = const . (D-12)

Eliminating instead T ∼ pV , the isentropic condition takes its most common form,

p V γ = const . (D-13)

Notice that the constants are diﬀerent in these three equations.

Example D.3.2: When the air in a bicycle pump is compressed from V0 to V1 (while you block the valve with your ﬁnger), the adiabatic law implies that γ γ p1V1 = p0V0 . For p0 = 1 atm and V1 = V0/2 we ﬁnd p1 = 2.6 atm. The temperature simultaneously rises about 100 degrees, but the hot air quickly becomes cold again during the backstroke. One may wonder why the fairly rapid compression stroke may be assumed to be reversible, but as long as the speed of the piston is much smaller than the velocity of sound, this is in fact a reasonable assumption. Conversely, we may conclude that the air expands with velocity close to the speed of sound when the wall is opened in example D.3.1.

Isothermal versus isentropic bulk modulus We have formerly seen that the bulk modulus of a strictly isothermal ideal gas with p = ρRT0/Mmol is equal to the pressure, µ ¶ ∂p KT = ρ = p . (D-14) ∂ρ T Here the index T (in the usual thermodynamic way of writing derivatives) signals that the temperature must be held constant while we differentiate. In terms of the mass density ρ = M/V , the isentropic condition may be written in any of three different ways (with three different constants),

p ρ−γ = const , T ρ1−γ = const ,T γ p1−γ = const . (D-15)

Using the first we find the isentropic bulk modulus of an ideal gas, µ ¶ ∂p KS = ρ = γ p , (D-16) ∂ρ S where the index S now signals that the entropy must be held constant. The distinction between the isothermal and isentropic bulk modulus is necessary in all materials, but for nearly incompressible liquids there is not a great difference between KS and KT . Among Isaac Newton’s great achievements was the first calculation of the speed of sound in air, using essentially the ideal gas law with constant temperature. His result did not agree with experiment, because normal sound waves oscillate so rapidly that compression and expansion are essentially isentropicp processes. In section 16.2 we shall find that the speed of sound is c = K/ρ, such that the √ ratio between the isentropic and isothermal sound velocities is cS /cT = γ. For air with γ ≈ 1.4 this amounts to an 18% error in the sound velocity. Much later in 1799, Laplace derived the correct value for the speed of sound.

D.4 Speciﬁc quantities

In classical thermodynamics we always think of a macroscopic volume of mat- ter with the same thermodynamic properties throughout the volume. Volume, mass, energy, entropy, and the specific heats are all extensive quantities, meaning that the amount of any such quantity in a composite system is the sum of the amounts in the parts. Pressure, temperature, and density are in contrast intensive quantities, that may not be added when a system is put together from its parts. In continuum physics, an intensive quantity becomes a field that may vary from place to place, whereas an extensive quantity becomes an integral over the density of the quantity. Since a material particle with a fixed number of molecules has a fixed mass (subject to the reservations set down in chapter 1), the natural field to introduce for an extensive quantity like the energy is the specific internal energy, u = dU/dM, which is the amount of energy per unit of mass in the neighborhood of a given point. The actual energy density becomes dU/dV = ρ u, and the total energy in a volume Z U = ρ u dV . (D-17) V The specific energy is an intensive quantity like temperature, pressure or density. Similarly, we define the specific heat as the local heat capacity per unit of mass. Since the heat capacities (D-6) of an ideal gas are directly proportional to the mass M = nMmol, the specific heats of an ideal gas become, 1 R γ R cV = , cp = . (D-18) γ − 1 Mmol γ − 1 Mmol They are constants which only depend on the properties of the gas. For air we have cV = 718 J/K/kg and cp = 1005 J/K/kg. From (D-2) we obtain after dividing by M,

u = cV T. (D-19) The specific energy of an ideal gas is the specific heat times the absolute temperature. Finally, we define the specific entropy, s = dS/dM, from which the total entropy may be calculated as an integral, Z S = ρ s dV . (D-20) V For an ideal gas, the specific entropy is obtained by dividing (D-10) by M = nMmol. It may be written in three different forms related by the ideal gas law, ¡ 1−γ ¢ s = cV log T ρ + const (D-21a) ¡ γ 1−γ ¢ = cV log T p + const (D-21b) ¡ −γ ¢ = cV log p ρ + const . (D-21c) Notice, however, that the constants are different in the three cases.

Problems P D.1 An ideal gas mixture contains n = i ni moles. (a) Show that the mixture obeys the equation of state (4-27) when the molar mass of the mixture is deﬁned as X Mmol = ciMmol,i . (D-22) i

where ci = ni/n of the molar fraction of the i-th component. (b) Show that the number of degrees of freedom of a mixture is X k = ciki , (D-23) i

where ki is the degrees of freedom of the i-th component.

D.2 a) Show that for a function Q = Q(T,V ) the diﬀerential takes the form dQ = AdT + BdV where ∂A/∂V = ∂B/∂T . b) Prove that this is not fulﬁlled for (D-7).